When originally writing the article, I tried to include the following fact as a corollary of the existential closedness of algebraically closed fields in the class of fields:
If V is a geometrically integral variety over a field K (i.e., the base change of V to the algebraic closure of K is integral), then every base change of V is integral.
The proof ended up getting a bit out of hand. I'm moving it to this talk page since it got a bit off topic for this article:
As another "application", let us prove:
Corollary: Let K be an algebraically closed field, and L be a field extending K. Let A be a K-algebra, possibly infinite dimensional. Suppose A is integral. Then so is .
(In algebro-geometric terms, this means that if a variety is geometrically integral, it remains integral under any base change.)
Proof: We will instead prove a slightly stronger statement: let V and W be K-vector space, and let be some K-bilinear pairing. Tensoring everything with L, we get
Suppose that whenever . Then has the same property.
To see this, write V as a directed limit of finite dimensional submodules
Let be the K-span of , so that induces a map for each , and the two maps
The functor is a left adjoint, so it preserves direct limits. Also, because K is a field, L is flat, so this functor preserves injections. Therefore,
So, if there exist two non-zero vectors with , then some already contains .
Replacing V with Vi and W with Wi, we may assume that V and W are finite dimensional. In fact, we may assume that they are Kn and Km for some n and m. Then is described by structure coefficients:
where the ei are standard basis vectors. The same structure coefficients are correct after tensoring with L.
Our assumption means that in K, the following system of equations cannot be solved:
in the variables . By existential closedness of K in L, the same system of equations cannot be solved in L, which is what we wanted to show. QED.
Will Johnson (talk) 07:15, November 15, 2013 (UTC) ::: ::: :::